Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a question on Sobolev space. This is one of exercises in Evans PDE textbook. Let $U=\{(x,y) | |x|<1, |y<1|\} \subset \mathbb{R}^2$. Define a function $u(x,y)$ by $$ u(x,y)=\begin{cases} 1-x & \text{if } x>0, \ |y|<x \\ 1+x & \text{if } x<0, \ |y|<-x \\ 1-y & \text{if } y>0, \ |x|<y \\ 1+y & \text{if } y<0, \ |x|<-y \end{cases} $$ I would like to know for which $p$ the function $u$ in $W^{1,p}(U)$.

It seems to me that weak derivatives are given by $u_x(x,y)=-1,1,0,0$ and $u_y(x,y)=0,0,-1,1$ in each region respectively, but then this question is too trivial. Could someone point out any mistake I made?

share|cite|improve this question
How did you find that? – Davide Giraudo Nov 4 '12 at 11:48

Your intuition is correct, but remember that you have to prove that the functions you give (which technically aren't functions, and only defined up to measure zero) are actually the weak derivatives of $u(x,y)$, i.e., calculate that $$\int_U v \varphi \, dx = - \int_U u \varphi_x \, dx$$ where $v$ is your candidate for $u_x$, etc, and $\varphi \in C_c^{\infty}(U)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.